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The point(s) on the parabola `y^2 = 4x` which are closest to the circle`x^2 + y^2 - 24y + 128 = 0` is/are |
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Answer» equation of circle `x^2+y^2-24y+128=0` its center=(0,12) equation of parabola `y^2=4x` where a=1 point P`(at^2,2at)=(t^2,2t)` normal at point (t) `y=-tx+at^3+2at` a=1 `y=-tx+t^3+2t` this equation should be normal for circle as well (0,12) should be satisfied `12=t^3+2t` 3+2t-12=0 (t-2)is a factor after division `(t-2)(t^2+2t+6)=0` but `(t^2+2t+6)!=0` so, (t-2)=0 t=2 putting this value in point P P(4,4) |
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